Gravitational Field Strength 

G.F.S at a point is defined as the force per unit mass acting on a
small mass placed at that point. 



Therefore, the units of G.F.S. are Nkg^{1}. 



However, Newton's second law of motion reminds us that force per unit
mass (that is, force divided by mass) is equivalent to acceleration. 

This means that G.F.S and acceleration due to gravity are, in fact,
different terms for the same quantity. 



G.F.S. at a point a distance r away from a point mass M 

Using Newton's law of gravitation and Newton's second law of motion (see
here) we have 



This shows that g (or G.F.S.) varies with the inverse square of the
distance r. 



So a graph of g against r is as shown below. 



G.F.S. at a point a distance r away from a the centre of a sphere of
radius R and mass M 



Outside the sphere, the same inverse square law applies. 

So, outside the sphere, the graph is as shown below. 





It can be shown that, inside the sphere, the G.F.S. is directly
proportional to the distance from the centre. 

So the complete graph is 



Note that, the maximum
value of g (of G.F.G.), is at the surface of the body. 



Variation of g (of G.F.S.) on a line joining the centres of two point masses 

Consider the following situation 



How does g vary if point p starts very close to mass m_{1} and moves to
being very close to mass m_{2}? 



When very close to m_{1} that mass "dominates" and so the curve is
pretty similar to the first graph, above (though inverted, due to our choice
of the positive sense being directed away from m_{1}) 



At some point in between the two masses we arrive at a distance where the
two fields have equal magnitude but opposite sense, so the G.F.S. equals zero
here. 

In the case shown here, this point is nearer to m_{2} which
means that the greater of the two masses is... answer below... 



We can
easily find the position of this point using: 



where D is the total distance between the centres of the two masses. 



As we approach m_{2} that mass starts to "dominate" the field and
so, again, we have a similar curve to the first graph in this region. 

Note that, as we approach either mass, the field strength approaches
infinity (since these are defined to be point masses having no
spacial distribution). 



In this case m_{1} > m_{2} 



Gravitational Potential, V 

The gravitational potential at a point in a gravitational field is defined
as the work done per unit mass bringing a small mass from
infinity to that point. 



The units of potential are therefore Jkg^{1} 



Why "from infinity" ? 

By saying "at infinity", we mean far enough
away to be able to assume that the force of gravity due to the body being
considered is effectively zero. 



Once we have reached this situation, we can move a body around
without doing any work against gravitational forces. 

For this reason we consider that a body "at infinity" has zero
potential. 

Rephrasing this statement: the zero of gravitational potential
is at infinity. 



To calculate a potential at a point we therefore use the
equation work done equals force multiplied by displacement. 

However, in this case the calculation
is complicated by the fact that the force varies as the mass is moved. 

We do know how the force varies with distance (Newton's law of
gravitation) 



where, in this case, m_{1} = M, the mass of the body "causing" the
field and m_{2} = m, the small mass we are moving. 

This allows us to find the following equation for work done*




This is the work done moving a mass m, so the work done per unit mass
is given by 





However, as stated above, the zero of potential is taken to be at infinity. 

Given the chance, a body always falls to its
lowest state of potential (energy) so we must arrange that the equation
for potential is such that, as r decreases, the potential also decreases.


We can do this by including a
negative sign in the above equation.


Then, as r decreases, V decreases (becomes a greater negative
quantity). 



Therefore the potential V at a point a distance r away from a point mass (or
from the centre of a uniformly distributed spherical mass) M is given by: 



This equation is useful in the calculation of a planet's escape velocity. 





*To obtain this equation, we imagine moving the mass one tiny
step at a time, each step being so small that we can assume that the
force does not vary appreciably. 

We then add up all these "infinitesimal" quantities of work to
find the total. 

The process is called integration. 
